$\dfrac{d}{dx}[-\text{ln}(x)+x^2]=$
Solution: Recall that ${\dfrac{d}{dx}[\text{ln}(x)]=\dfrac1x}$ and ${\dfrac{d}{dx}[x^n]=nx^{n-1}}$. $\begin{aligned} &\phantom{=}\dfrac{d}{dx}[-\text{ln}(x)+x^2] \\\\ &=-1\cdot{\dfrac{d}{dx}[\text{ln}(x)]}+{\dfrac{d}{dx}[x^2]} \\\\ &=-1\cdot{\dfrac1x}+({2x^1}) \\\\ &=-\dfrac1x+2x \end{aligned}$ In conclusion, $\dfrac{d}{dx}[-\text{ln}(x)+x^2]=-\dfrac1x+2x$